No, sector $-t<x<t$ has not 2 values (it would be if $u=\left\{\begin{aligned} &1 &&x<0,\\ -&1 &&x>0\end{aligned}\right.$). So far it has NO values at all.

Now hint:

a) consider $u_s (x,t)= s^l u(s x, s^k t)$ and prove that if $u$ satisfies original problem then $u_s$ satisfies it for all $s>0$ iff $k=1$, $l=0$.

b) So, $u_s(x,t)= u(sx, st)$ satisfies it and we are interested in self-similar solution $u(x,t)=u(sx,st)$ for all $s>0$. Plugging $s=t^{-1}$ we arrive to $u(x,t)=v (xt^{-1})$ (with $v(y)=u(y, 1)$.

c) Plugging $u(x,t)=v (xt^{-1})$ into original equation we have an ODE. Which?

d) Find continuous solution of this ODE such that $v(y)=-1$ as $y<-1$ and $v(y)=1$ as $y>1$ (Think why).

e) Plug into $u$